A bag contains $2$ red marbles, $2$ green marbles, and $4$ blue marbles. If we choose a marble, then another marble without putting the first one back in the bag, what is the probability that the first marble will be red and the second will be green?
Solution: The probability of event A happening, then event B, is the probability of event A happening times the probability of event B happening given that event A already happened. In this case, event A is picking a red marble and leaving it out. Event B is picking a green marble. Let's take the events one at at time. What is the probability that the first marble chosen will be red? There are $2$ red marbles, and $8$ total, so the probability we will pick a red marble is $\dfrac{2} {8}$. After we take out the first marble, we don't put it back in, so there are only $7$ marbles left. Since the first marble was red, there are still $2$ green marbles left. So, the probability of picking a green marble after taking out a red marble is $\dfrac{2} {7}$. Therefore, the probability of picking a red marble, then a green marble is $\dfrac{2}{8} \cdot \dfrac{2}{7} = \dfrac{4}{56} = \dfrac{1}{14}$